What is the tangent of y=sqrt(x),(1,1) ?

The tangent of y=sqrt(x),(1,1) is y= 1/2 x+1/2

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tangent of y=sqrt(x),(1,1)

tangent of

y = x, (1, 1)

y = \frac{1}{2} x + \frac{1}{2}

Solution steps

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

(1, 1) : m = \frac{1}{2}

Find the line with slope m=

\frac{1}{2}

and passing through

(1, 1) : y = \frac{1}{2} x + \frac{1}{2}

y = \frac{1}{2} x + \frac{1}{2}

\frac{x}{x + 3}

n = 1 \sum \infty (\frac{3}{5})^{n}

\frac{d}{d t} (2 sin (t) - sin (2 t))

\int (6 x - 1) d x

5 x, 9 x^{2}

What is the tangent of y=sqrt(x),(1,1) ?
The tangent of y=sqrt(x),(1,1) is y= 1/2 x+1/2